Let $P(x)=4x^2+6x+4$ and $Q(x) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$. 
Let $P(x)=4x^2+6x+4$ and $Q(y) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$.   

I know this question has been asked earlier here but the answer given states "Note that we have...". I want to know how we can arrive at that statement mathematically instead of trial and error method, because I wasn't able to get that while trying to solve it.    
Any other way of solving this question would also be helpful.
 A: In the solution of the problem you linked to it was shown that completing the square gives the result that
$$  P(x)Q(y)=\left[\left(2x+\frac{3}{2}\right)^2+\frac{7}{4}\right]\cdot\left[(2y-3)^2+16\right]$$
If you multiply out those terms you get three terms which are non-negative plus a constant term of $28$. Therefore the product must be no smaller than $28$. In order for the product to exactly equal $28$ each of the non-negative terms must equal $0$. Thus
$2x+\frac{3}{2}=0$ and $2y-3=0$. Therefore we know that $x=-\frac{3}{4}$ and $y=\frac{3}{2}$. Then finding $11y-26x$ is easy.
A: It can be done by completing the square method,
$P(x)=4x^2+6x+4$
Let P(x) = 0.
Then,
$4x^2+6x+4 = 0$
First divide equation by 4,
So we have,
$x^2+\frac64x+1 = 0$
$x^2+\frac32x = -1$
The term of x is positive so compare it with,
$a^2 + 2ab + b^2 = (a + b)^2$
So, a = x, 
And 2ab = $\frac32x$
2b = $\frac32$
b = $\frac34$
Then $b^2 = \frac{9}{16}$
So above equation become,
$x^2+\frac32x+\frac9{16}= -1+\frac{9}{16}$
$(x+\frac34)^2 = \frac{-7}{16}$
$(x+\frac34)^2 + \frac{7}{16} = 0$
In Q(x) also divide the equation by 4 and then compare with 
$a^2 - 2ab + b^2 = (a - b)^2$ as term containing x is negative.
